# Rearranging an equation

• Nov 11th 2007, 11:24 AM
Topin
Rearranging an equation
Hi,
I currently have the equation:

γ tan(α) = sin(α') / (cos(α') + β)

I was using α as the subject however in order to do my calculations I need α' to be the subject.
Does anyone know how I could rearrange this to that I have one case of α' on its own?
Thanks
• Nov 11th 2007, 12:01 PM
topsquark
Quote:

Originally Posted by Topin
Hi,
I currently have the equation:

γ tan(α) = sin(α') / (cos(α') + β)

I was using α as the subject however in order to do my calculations I need α' to be the subject.
Does anyone know how I could rearrange this to that I have one case of α' on its own?
Thanks

Ummm....

In terms of symbols, is this supposed to be
$\displaystyle \gamma ~ tan(\alpha) = \frac{sin(\alpha ^{\prime})}{cos(\alpha ^{\prime} ) + \beta}$

Anyway:
$\displaystyle \gamma ~ tan(\alpha) ~ ( cos(\alpha ^{\prime} ) + \beta ) = sin(\alpha ^{\prime})$

$\displaystyle \gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ) = sin(\alpha ^{\prime})$

Because we don't know which quadrant $\displaystyle \alpha ^{\prime}$ is in we need to do the following:
$\displaystyle (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = sin^2(\alpha ^{\prime})$

And $\displaystyle sin^2( \alpha ^{\prime} ) = 1 - cos^2( \alpha ^{\prime} )$, so
$\displaystyle (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )$

When you simplify this, you will have a quadratic equation in $\displaystyle cos( \alpha ^{\prime} )$ which you can do the quadratic equation on.

-Dan
• Nov 11th 2007, 01:52 PM
Topin
Quote:

$\displaystyle (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )$
I think there was a missing bracket but I tried to rearrange this but I couldn't get all of the $\displaystyle cos(\alpha ^{\prime} )$ together in a way that I could then simplify. The last line of my working before I seemed to double back on myself was:
$\displaystyle cos^2(\alpha ^{\prime} ) + 2 cos(\alpha ^{\prime} ) \beta + \beta^2 = \frac{1 - cos^2(\alpha ^{\prime} )} {\gamma^2 tan^2 (\alpha)}$
I was wondering where I could procede from here. Thanks for your help so far though.
• Nov 11th 2007, 09:10 PM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta )^2 = 1 - cos^2( \alpha ^{\prime} )$

$\displaystyle \gamma ^2 ~ tan^2(\alpha) ~ cos^2(\alpha ^{\prime} ) + 2 \gamma~tan(\alpha)~cos(\alpha ^{\prime} ) \cdot \gamma \beta + \gamma^2 \beta^2 = 1 - cos^2( \alpha ^{\prime} )$

$\displaystyle (\gamma ^2 ~ tan^2(\alpha) + 1) cos^2(\alpha ^{\prime} ) + 2 \gamma^2 \beta~tan(\alpha)~cos(\alpha ^{\prime} ) + \gamma^2 \beta^2 - 1 = 0$

Set $\displaystyle y = cos(\alpha ^{\prime} )$.

Then we have
$\displaystyle (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + 2 \gamma^2 \beta~tan(\alpha)~y + (\gamma^2 \beta^2 - 1) = 0$
which is a quadratic in y.

-Dan
• Nov 12th 2007, 01:49 AM
Topin
Finally done it now. I found that in my example it came to
$\displaystyle (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0$

Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for :D
Thankyou very much for your help - would never have managed this on my own!
• Nov 12th 2007, 08:52 AM
topsquark
Quote:

Originally Posted by Topin
Finally done it now. I found that in my example it came to
$\displaystyle (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0$

Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for :D
Thankyou very much for your help - would never have managed this on my own!

It's a useful trick. (Nod)

-Dan